1. Revision 2. Revision pv 3. - note that there are other equivalent formulae! 1 pv A x A 1 x:n A 1

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1 Tutorial 1 1. Revision 2. Revision pv 3. - note that there are other equivalent formulae! 1 pv A x A 1 x:n A 1 x:n a x a x:n n a x 5. K x = int[t x ] - or, as an approximation: T x K x Male: 0.993, Female: 0.994, Male: 0.007, Female: 0.006, l x+n l x+n+m l x or ( n p x )( m q x+n ) 9. Note for this question the interest rates should have been 0%,5% and 10.25% as noted in the tutorial A 100 = 1;0.894;0.803 (males), variance = A 100 = 1;0.881;0.781 (females), variance =

9617 7. Male: 0.007, 0.063 Female: 0.006, 0.0383 8. l x+n l x+n+m l x or ( n p x )( m q x+n ) 9.

2 Tutorial 2 1. Proof - see next page 2. Proof Proof - write down a formula for e x and e x+1, then multiply the second by p x and add p x 5. µ x = 1 l x d dx l x l x µ x = d dx l x R 0 l x+t µ x+t dt = R 0 6. Proof ; ; dl x+t dx+t dt = [l x+t] 0 = [l l x ] = l x ; if your first answer was note the UDD ( s q x sq x ) only applies for integer x and s (0,1], you first need to derive another approx is needed for non-integer x as we did in the tutorial ; l x = k( e x )Ax 11. $61, $61, , the second part of the question was scrapped...

3 Proof: tutorial 2 question 1 f x (t) = d dt F x(t) = d dt (P[T x t]) 1 = lim h 0 + h (P[T x t + h] P[T x t]) = lim (P[T x +t + h T > x] P[T x +t T > x]) h 1 x +t + h] P[T x] [T x +t] P[T x] = lim (P[T )) h 0 + h P[T > x] P[T > x] 1 1 = lim (P[T x +t + h] P[T x +t]) h 0 + h S(x) 1 S(x +t) P[T x +t + h] P[T x +t] = lim ( ) h 0 + h S(x) S(x +t) h = S x (t) lim h = S x (t)µ x+t = t p x µ x+t P[T x +t + h T > x +t] h

P[T > x] P[T > x] 1 1 = lim (P[T x +t + h] P[T x +t]) h 0 + h S(x) 1 S(x +t) P[T x +t + h] P[T x +t] = lim

4 Tutorial A x:n (A x:n ) 2 where 2 A x:n denotes A x:n calculated at i = (1 + i) Proof - do by splitting A 1 x:n into A x - n A x. Needs to be done from first principles - so define all three of these as the expected value of random variables and split the values by K x cov[x,y ] = 2 A 1 x:n (A x)(a 1 x:n ) Var[J] = 2 n A x ( n Ax) 2 where J is the random variable denoting the present value of the deferred assurance as defined in the first half of the question. Note that it is easier here to calculate Var[J] from first principles but that was not what the question asked ä x:n = n 1 j=0 j p x v j Var[ä min[kx +1,n] ] = 1 d 2 [ 2 A x:n (A x:n ) 2 ] ä x:n = a x:n + 1 v n np x Var[ä min[kx +1,n] ] = 1 d 2 [ 2 A x:n (A x:n ) 2 ] 4. Denote the present value of the payments under the deferred annuity due by Y. Y = 0 if K x n Y = v n a Kx n+1 if K x > n Therefore, in one term, Y = v n ä max[kx n+1,0]... Var[v n ä max[kx n+1,0] ] = 1 d 2 [E[(v max[k x+1,n] ) 2 ] (E[v max[k x+1,n] ) 2 ]... Var[v n ä max[kx n+1,0] ] = v2n d 2 [ n q x + n p x 2 A x+n ( n q x + n p x A x+n ) 2 ] For the deferred annuity; it is not possible to simplify much past 1 i 2 Var[v max[k x,n] ] 5. R 0 v t t p x dt 1 [ 2 A δ 2 x ( A x ) 2 ] 6. 3,4,5 year select periods: l 46 l 50 l [41]+2 6 year select: l [41]+5 l 50 l [41]+2 7. Expected values: 4%: 45,640 6%: 32,692 4% select: 45,510 6% select: 32,533 4% payable immediately on death: two approximations possible; one is claims accelaration; the other you end up multiplying by i δ

Ax) 2 where J is the random variable denoting the present value of the deferred assurance as defined in the first half of the question.

5 46,544 or 46,547 depending which one you use (rounded to the nearest ). 6% payable immediately on death: 33,658 or 33,663 Variances: 4%: 289,290,000 6%: 341,033, ä x:n 1 2 (1 vn np x )

6 Tutorial 4 1. N x+1 N x+n see Tut 3!, N x+m+1 3. Proofs - except: ä (m) x 4. Proof 5. t V x:n = 1 äx+t:n t ä x:n, N x+m+1 N x+m+n+1, N x ä x m 1 2m, M x M x+n, M x M x+n ++n, M x+m 6. TYPO in question: should be x : n in the last subscript Answer: because the reserve required for an endowment assurance is the reserve required for an endowment, plus the reserve required for an assurance; as an endowment assurance is both an endowment and an assurance... (if the policyholder dies, we need to have the reserve for an assurance, if he survives, we need to have the reserve for a pure endowment).

TYPO in question: should be x : n in the last subscript Answer: because the reserve required for an endowment assurance is the reserve required for an

7 Tutorial - formulae and tables 1. Note for this question that look up in the tables means that you should use available values and simplifying formulae (such as ä x = a x + 1) to arrive at your answer, not that the values are necessarily published somewhere in your yellow books. Not all variances are given, some more should be solvable. a 50:15 = commutation functions a 50:15 = from published figs in the tables Variance: a 50 = a 50 = Variance: a 50:15 = a 50:15 = ä 50 = ä 50 = A 1 50:15 = A 1 50:15 = A 50:15 = A 50:15 = A 50 = A 50 = %: 45,640, variance 289,290,400 6%: 32,692, variance 341,033,136 select 4%: 45,510, variance 283,539,900 select 6%: 32,533, variance 332,303,911 payable immediately on death: 4%: 46,544 payable immediately on death: 6%: 33, UDD: CFM: A 71:3 = (4%) A 71:3 = (6%) ä 63:20 = (4%) ä 63:20 = (6%) note that the question should have said over each of the next two years to avoid any potential confusion 6. A 1 x:n = Variance:

necessarily published somewhere in your yellow books. Not all variances are given, some more should be solvable. a 50:15 = 10.75750214 commutation functions a 50:15 = 10.

8 Tutorial 5 1. Proof Reserves - see spreadsheet printout Reserves - see spreadsheet printout 4. Proof 5. (DA) 1 x:n = (n + 1)A1 x:n (IA)1 x:n ( 13 24äx a x( 1 1+ j )) 7. 2, (Īā) x = R 0 tv t t p x dt (Iā) x = t=0 t ā x

101686157 Reserves - see spreadsheet printout 4. Proof 5.

9 Sheet1 net premium reserve: (question 2) premium: prospective retrospective annuity assurance total annuity assurance total net premium reserve: (question 3) premium: prospective retrospective annuity pure end total annuity pure end total Page 1

000417 0.000071 8.39 0 0.000071 8 net premium reserve: (question 3) premium: prospective retrospective 0.101835550 annuity pure end total annuity pure end total 1 6.15 0.732932 0.106454 1.05 0 0.

10 Tutorial 6 1. t V x:n = v[q x+t + p x+t ( t+1 V x:n )] P x:n Premiums, retrospective and prospective reserves all need to be calculated on the same basis. 2. Expected Death Strain: 1,380,000 Mortality Profit: - 12,920,000 Revised EDS: 1,378,620 Revised Mortality Profit: - 12,907, Proof 4. EDS first year: Mortality profit: EDS 10 years in: EDS year beginning age 69: ADS year beginning age 69: Full solution to this is available here: data/assets/pdf_file/0007/141973/fandi_ct5_200809_report.pdf (question 12)

Expected Death Strain: 1,380,000 Mortality Profit: - 12,920,000 Revised EDS: 1,378,620 Revised Mortality Profit: - 12,907,080 3. Proof 4.

Math 630 Problem Set 2

Math 630 Problem Set 2 Math 63 Problem Set 2 1. AN n-year term insurance payable at the moment of death has an actuarial present value (i.e. EPV) of.572. Given µ x+t =.7 and δ =.5, find n. (Answer: 11) 2. Given: Ā 1 x:n =.4275,